If $$y = {\left( {x + \sqrt {1 + {x^2}} } \right)^n},$$ then $$\left( {1 + {x^2}} \right){{{d^2}y} \over {d{x^2}}} + x{{dy} \over {dx}}$$ is

$${\cot ^{ - 1}}\left( {\sqrt {\cos \alpha } } \right) - {\tan ^{ - 1}}\left( {\sqrt {\cos \alpha } } \right) = x,$$ then sin x is equal to :

The maximum distance from origin of a point on the curve
$$x = a\sin t - b\sin \left( {{{at} \over b}} \right)$$
$$y = a\cos t - b\cos \left( {{{at} \over b}} \right),$$ both $$a,b > 0$$ is

If $$a>0$$ and discriminant of $$\,a{x^2} + 2bx + c$$ is $$-ve$$, then
$$\left| {\matrix{ a & b & {ax + b} \cr b & c & {bx + c} \cr {ax + b} & {bx + c} & 0 \cr } } \right|$$ is equal to